Question -
Answer -
Volume of the solidsphere = (4/3)╧Аr3
Volume of twenty sevensolid sphere = 27├Ч(4/3)╧Аr3┬а= 36 ╧А r3
(i) New solid ironsphere radius = rтАЩ
Volume of this newsphere = (4/3)╧А(rтАЩ)3
(4/3)╧А(rтАЩ)3┬а=36 ╧А r3
(rтАЩ)3┬а=27r3
rтАЩ= 3r
Radius of new spherewill be 3r (thrice the radius of original sphere)
(ii) Surface area ofiron sphere of radius r, S =4╧Аr2
Surface area of ironsphere of radius rтАЩ= 4╧А (rтАЩ)2
Now
S/SтАЩ = (4╧Аr2)/(4╧А (rтАЩ)2)
S/SтАЩ = r2/(3rтАЩ)2┬а=1/9
The ratio of S and SтАЩ is 1:9.